Straightforward question: student must find the general solution to a second order constant coefficient ODE. Uses custom marking algorithm to check that both roots appear and that the solution is in the correct form (e.g. two arbitrary constants are present). Arbitrary constants can be any non space-separated string of characters. The algorithm also allows for the use of $e^x$ rather than $\exp(x)$.

Unit tests are also included, to check whether the algorithm accurately marks when the solution is correct; when it's correct but deviates from the 'answer'; when one or more roots is incorrect; or when the roots are correct but constants of integration have been forgotten.

This is the question for week 3 of the MA100 course at the LSE. It looks at material from chapters 5 and 6. The following describes how two polynomials were defined in the question. This may be helpful for anyone who needs to edit this question.

In part a we have a polynomial. We wanted it to have two stationary points. To create the polynomial we first created the two stationary points as variables, called StationaryPoint1 and StationaryPoint2 which we will simply write as s1 ans s2 here. s2 was defined to be larger than s1. This means that the derivative of our polynomial must be of the form a(x-s1)(x-s2) for some constant a. The constant "a" is a variable called PolynomialScalarMult, and it is defined to be a multiple of 6 so that when we integrate the derivative a(x-s1)(x-s2) we only have integer coefficients. Its possible values include positive and negative values, so that the first stationary point is not always a max (and the second always a min). Finally, we have a variable called ConstantTerm which is the constant term that we take when we integrate the derivative derivative a(x-s1)(x-s2). Hence, we can now create a randomised polynomial with integers coefficients, for which the stationary points are s1 and s2; namely (the integral of a(x-s1)(x-s2)) plus ConstantTerm.

In part e we created a more complicated polynomial. It is defined as -2x^3 + 3(s1 + s2)x^2 -(6*s1*s2) x + YIntercept on the domain [0,35]. One can easily calculate that the stationary points of this polynomials are s1 and s2. Furthermore, they are chosen so that both are in the domain and so that s1 is smaller than s2. This means that s1 is a min and s2 is a max. Hence, the maximum point of the function will occur either at 0 or s2 (The function is descreasing after s2). Furthermore, one can see that when we evaluate the function at s2 we get (s2)^2 (s2 -3*s1) + YIntercept. In particular, this is larger than YIntercept if s2 > 3 *s1, and smaller otherwise. Possible values of s2 include values which are larger than 3*s1 and values which are smaller than 3*s1. Hence, the max of the function maybe be at 0 or at s2, dependent on s2. This gives the question a good amount of randomisation.

Straightforward question: student must find the general solution to a second order constant coefficient ODE. Uses custom marking algorithm to check that both roots appear and that the solution is in the correct form (e.g. two arbitrary constants are present). Arbitrary constants can be any non space-separated string of characters. The algorithm also allows for the use of $e^x$ rather than $\exp(x)$.

Unit tests are also included, to check whether the algorithm accurately marks when the solution is correct; when it's correct but deviates from the 'answer'; when one or more roots is incorrect; or when the roots are correct but constants of integration have been forgotten.

This uses an embedded Geogebra graph of a polar function with random coefficients set by NUMBAS.

No description given

Equating coefficients of a polynomial. Basic ones that don't require simultaneous equations.

Very good feedback and corresponds to instance of randomisation

Equating coefficients of a polynomial. Basic ones that don't require simultaneous equations.

No description given

More work on differentiation with fractional coefficients.

Solve: $\displaystyle \frac{d^2y}{dx^2}+2a\frac{dy}{dx}+(a^2+b^2)y=0,\;y(0)=1$ and $y'(0)=c$. 

Solve: $\displaystyle \frac{d^2y}{dx^2}+2a\frac{dy}{dx}+(a^2+b^2)y=0,\;y(0)=1$ and $y'(0)=c$. 

Find the solution of a constant coefficient second order ordinary differential equation of the form $ay''+by'+cy=0$.

This question tests the students ability to factorise simple quadratic equations (where the coefficient of the x^2 term is 1) and use the factorised equation to solve the equation when it is equal to 0.

Factorise three quadratic equations of the form $x^2+bx+c$.

The first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.

Use a series of difference tables to determine the coefficients of a degree three polynomial function from a set of data values, equally spaced over the x-axis.

Find the leading coefficient of a degree 3 polynomial using a difference table.

Solve: $\displaystyle \frac{d^2y}{dx^2}+2a\frac{dy}{dx}+(a^2+b^2)y=0,\;y(0)=1$ and $y'(0)=c$. 

A copy of Amy Chadwicks block question, with adaptions to provide a diagram for the student.

A block of given mass is sliding down the plane, with given acceleration. Find the normal reaction force, the coefficient of friction, and the distance travelled before reaching a given speed.

Factorise a quadratic equation where the coefficient of the $x^2$ term is greater than 1 and then write down the roots of the equation

Algebra:  Quadratic Factorisation.

              Coefficient of squared term is 1.

Algebra:  Quadratic Factorisation.

              Coefficient of squared term is 1.

No description given

Equating coefficients of a polynomial. Basic ones that don't require simultaneous equations.

Factorise three quadratic equations of the form $x^2+bx+c$.

The first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.

Algebra:  Quadratic Factorisation.

              Coefficient of squared term is 1.

Factorise three quadratic equations of the form $x^2+bx+c$.

The first has two negative roots, the second has one negative and one positive, and the third is the difference of two squares.

Putting a pair of linear equations into matrix notation and then solving by finding the inverse of the coefficient matrix. 

Find a regression equation given 12 months data on temperature and sales of a drink. 

A block of given mass is sliding down the plane, with given acceleration. Find the normal reaction force, the coefficient of friction, and the distance travelled before reaching a given speed.